Statement 1: `f(x)=(x^3)/3+(a x^2)/2` + x+5 has positive point of maxima for `a<-2.` Statement 2: `x^2+a x+1=0` has both roots positive for `a<2.`

Let f(x) =x^(3) +3(a-7)x^(2)+3(a^(2) -9) x-1. If f(x) has positive point of maxima then find possible values of 'a'

f(x)=2x^(3)-9x^(2)+12x+5 has maximum at the point

f(x)=2x^(3)+3x^(2)-12x+7 has maximum at the point

f(x)=2x^(3)-3x^(2)-12x+12 has minimum at the point

Let f(x) =x^(3) =x^(2) -x-4 (i) find the possible points of maxima // minima of f(x) x in R (ii) Find the number of critical points of f(x) x in [1,3] (iii) Discuss absolute (global) maxima // minima value of f(x) for x in [-2,2] (iv) Prove that for x in (1,3) the function does not has a Global maximum.

The least positive integral value of lambda for which f(x)=(3x^(3))/(2)+(lambda x^(2))/(3)+x+7 has a point of maxima

f(x)=2x^(3)+3x^(2)-12x+5 has maximum value at x^(=) …

Show that the function f(x)=4x^(3)-18x^(2)+27x-7 has neither maxima nor minima.